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JulsSmile [24]
4 years ago
7

Scott Harris can invest $7,000 in a 1-year CD that earns interest at an annual rate of 4 percent compounded monthly. The amount

per $1.00 is 1.040742. He can also invest $7,000 in a 1-year CD at annual rate of 4 percent compounded quarterly. The amount per $1.00 is 1.040604. What is the difference in the amount of interest earned for each investment?
A) $0.96
B) $0.81
C) $0.87
D) $0.88
Mathematics
1 answer:
Tom [10]4 years ago
8 0

Answer:

option A  is answer.

Step-by-step explanation:

Amount of interest earned for compounded monthly  for $1 = 1.040742

Amount of interest earned for compounded quarterly for $1= 1.040604

Difference for $1  =   1.040742 - 1.40604

                 =0.000138

Difference for $7000 = $7000x0.000138

                                    = 0.966

                                      or $0.96

               

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Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

                                 P.Q. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of people who disagree with the statement = \frac{1812}{2625} = 0.69

           n = sample of respondents = 2625

           p = population proportion of people who disagree with statement

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

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P(-2.58 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.58) = 0.99

P( -2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [ \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ]

= [ 0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } , 0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } } ]

 = [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

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Answer:

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